Post on 03-Oct-2020
THERMOCHEMISTRYCP Unit 9
Chapter 17
Thermochemistry 17.1p Thermochemistry is the study of energy
changes (HEAT) that occur during chemical reactions and changes in state.
ENTHALPY 17.2p Enthalpy = a type of chemical energy, sometimes
referred to as “heat content”, ΔH (the heat of reaction for a chemical reaction)
p endothermic reactions (feels cold):n q = ΔH > 0 (positive values)
p exothermic reactions (feels hot):n q (heat) = ΔH (enthaply, heat of rxn) < 0 (negative
values)
Endo vs. Exo-Endothermic reactions:
absorbs heat from surroundings (+). n If you touch an
endothermic reaction it feels COLD
Exothermic reactions: release heat to the surroundings (-)n If you touch an
exothermic reaction it feels HOT
Magnitude of Heat Flow• Units of heat energy:
• 1 kcal = 1,000 cal = 1 Cal (nutritional)• 1 kJ = 1,000 J• 1 calorie = 4.184 J• 1 kcal = 4.184 kJ
Thermochemical Equations• A chemical equation that shows the
enthalpy (DH) is a thermochemical equation.
Rule #1The magnitude (value) of DH is directly proportional to the amount of reactant or product.
H2 + Cl2 ® 2HCl DH = - 185 kJ
* meaning there are 185 kJ of energy RELEASED for every:
1 mol H21 mol Cl22 moles HCl
Rules of ThermochemistryExample 1:H2 + Cl2 ® 2HCl DH = - 185 kJ
Calculate DH when 2.00 moles of Cl2reacts.
Rules of ThermochemistryExample 2: Methanol burns to produce carbon dioxide and water:
2CH3OH + 3O2 ® 2CO2 + 4H2O DH = - 1454 kJ
What mass of methanol is needed to produce 1820 kJ?
Rule #2DH for a reaction is equal in the magnitude but opposite in sign to DH for the reverse reaction.
(If 6.00 kJ of heat absorbed when a mole of ice melts, then 6.00 kJ of heat is given offwhen 1.00 mol of liquid water freezes)
Rules of ThermochemistryExample 1:Given: H2 + ½O2 ® H2O DH = -285.8 kJ
Reverse: H2O è H2 + ½O2 DH = +285.8 kJ
Example 2CaCO3 (s) è CaO (s) + CO2 (g) DH = 178 kJ
What is the DH for the REVERSE RXN?
CaO (s) + CO2 (g) è CaCO3 (s) DH = ?
Alternate form of thermochem. eq.p Putting the heat content of a reaction
INTO the actual thermochemical eq.p H2 + ½O2 ® H2O DH = -285.8 kJ
Exothermic
(DH is negative)
Heat is RELEASED as a PRODUCT
The alternate form is this:
H2 + ½O2 ® H2O +285.8 kJ
EX:2 NaHCO3 + 129 kJ è Na2CO3 + H2O + CO2
Put in the alternate form
The alternate form is this:2 NaHCO3 è Na2CO3 + H2O + CO2 DH = + 129 kJ
Put the following in alternate form1. H2 + Cl2 è 2 HCl DH = -185 kJ
2. 2 Mg + O2 è 2 MgO + 72.3 kJ
3. 2 HgO è 2 Hg + O2 DH = 181.66 kJ
Enthalpies of Formationenthalpy
change (delta)
standard conditions
formation
°D fH
Enthalpies of Formationpusually exothermicp see table for DHf
° value (Table A3)penthalpy of formation of an element
in its stable state = 0p these can be used to calculate DH°
for a reaction
Standard Enthalpy ChangeStandard enthalpy change, DH°, for a given thermochemical equation is = to the sum of the standard enthalpies of formation of the product – the standard enthalpies of formation of the reactants.
) H(-) H(H reactantsfproductsfrxn°°° DD=D SS
sum of (sigma)
Standard Enthalpy Changep elements in their standard states can be omitted:
2 Al(s) + Fe2O3(s) ® 2 Fe(s) + Al2O3(s)
ΔHrxn = S(ΔHf°products) - S(ΔHf
°reactants)
ΔHrxn = ΔHf°Al2O3 - ΔHf
°Fe2O3
ΔHrxn = (-1676.0 kJ) – (-822.1 kJ)
ΔHrxn = -853.9 kJ
Standard Enthalpy Changep the coefficient of the products and reactants in
the thermochemical equation must be taken into account:
O2(g) + 2SO2 (g) ® 2SO3 (g)
ΔHrxn = S(ΔHf°products) - S(ΔHf
°reactants)
ΔHrxn = 2ΔHf°SO3 - 2ΔHf
°SO2
ΔHrxn = 2(-395.7 kJ) – 2(-296.8 kJ)
ΔHrxn = -197.8 kJ
Standard Enthalpy Change - Calculate the standard heat for formation of benzene, C6H6, given the following thermochemical equation:
C6H6(l) + 15/2 O2(g) ® 6CO2(g) + 3H2O(l) DH° = -3267.4 kJ
-3267.4kJ = [6(-393.5kJ)+3(-285.8kJ)]–X
X = +49.0 kJ
-3267.4kJ = -3218.4–X
-49.0kJ = – X
X Total-393.5kJ -285.8kJ
Don’t forget the coefficents!